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Question: How do you solve\(\dfrac{{x - 7}}{{x - 1}} < 0\)?...

How do you solvex−7x−1<0\dfrac{{x - 7}}{{x - 1}} < 0?

Explanation

Solution

This question is related to the topic of solving one variable linear inequalities. We solve this question similarly as we solve the one variable linear equalities. To solve the linear inequalities of one variable, we will first move all variables term to one side of the equation and all constant terms of the equation to the other side of the equation by performing necessary operations.

Complete step by step solution:
Let us try to solve this question in which we need to find the value of xx such that the expression x−7x−1\dfrac{{x - 7}}{{x - 1}}always less than00. We will solve for linear inequalities, similarly as linear equalities.

The only difference is that we get an exact value of xxwhich satisfies the linear equality whereas we get a set of values of xxwhich satisfy the given linear inequality.

Now let us solve for this inequality, we have
x−7x−1<0\dfrac{{x - 7}}{{x - 1}} < 0

We cannot here perform the cross multiplication, because the denominator consists of variable terms of which value we are finding. So we write the numerator part of the expression such that we can perform separation of constant and variable term.

We can write above equation as follows
x−1−6x−1<0\dfrac{{x - 1 - 6}}{{x - 1}} < 0

Now taking x−1x - 1as single we can write our equation as follows,
x−7x−1= x−1x−1−6x−1<0\dfrac{{x - 7}}{{x - 1}} = \,\dfrac{{x - 1}}{{x - 1}} - \dfrac{6}{{x - 1}} < 0

So it becomes,
1−6x−1<01 - \dfrac{6}{{x - 1}} < 0

Now taking 11to R.H.S of the equation we get,
−6x−1<\-1\dfrac{{ - 6}}{{x - 1}} < \- 1

Now doing cross multiplication, we get
−6<\-(x−1)- 6 < \- (x - 1)
−6<\-x+1- 6 < \- x + 1
x<7x < 7

Hence the value of xxfor the inequality x−7x−1<0\dfrac{{x - 7}}{{x - 1}} < 0is any value of x<7x < 7will satisfy it.

Note: The questions in which we are given a linear inequality of one variable and we have to solve for the value of the variable very easily. While solving for a linear inequality of one variable you just have to be clear about the sign of constant and variable terms where most students make mistakes.